Can this be proved?

Question

Prove that
(a-b)^(n) + nk(a-b) = v^(n)
doesn't exist.Here a,b,k & v are integers & n is a prime power greater then 2.

Answer 1

This can be proved by contradiction.
Assume that the following statement is true:
(a - b)^n + n*k*(a - b) = v^n, where a, b, k and v are integers and n is a prime number > 2.

Now, we need to find a counter example, where the statement violates.

Let n = 3, k = 5, a = 4 and b = 2. Then,
(a - b)^n + n*k*(a - b)
= (4 - 2)^3 + 3*5*(4 - 2)
= 2^3 + 15*2
= 8 + 30
= 38

Now, let v^n = 38.
We want to find the value of v, where n = 3.
Using the logarithms,
log(v^3) = log(38)
3*log(v) = log(38)
log(v) = log(38)/3
Now, taking the exponentials,
v = exp(log(38)/3) = exp(log(38))*exp(1/3) = 38*exp(1/3).

[
If you haven't done logarithms before, then this might help:
1^3 = 3; 2^3 = 8; 3^3 = 27; 4^3 = 64.
Hence, there is no integer v such that v^n = 38.
]
So, we have shown that v is not an integer for the chosen values of a, b, n and k. This contradicts the statement that (a - b)^n + n*k*(a - b) = v^n, where a, b, k and v are integers and n is a prime number > 2 is true.

Therefore, the given statement is true.
i.e. (a - b)^n + n*k*(a - b) = v^n does not exist; where a, b, k and v are integers and n is a prime number > 2.

Hope this answers your question.

Answer 2

But it does: let a = 2, b = 1 (so a - b = 1), n = 3, v = 4. Then 1 + 3k*1 = 64, k = 21. Are you sure you copied this right?

Answer 3

5 variables, 1 equation. It's pretty unlikely you'll get an easily proved impossibility that way.